Linear Ordinary Differential Equations of First and Second Order - 6 | Physics for IIT JAM, UGC - NET, CSIR NET PDF Download

Mechanical Vibrations 

In this chapter we study some applications of the IVP

ay′′ + by′ + cy = g(t), y(0) = y0, y′(0) =

The spring-mass system: See figure below.

Figure (A): a spring in rest, with length l.

Figure (B): we put a mass m on the spring, and the spring is stretched. We call length L the elongation.

Figure (C): The spring-mass system is set in motion by stretch/squueze it extra, with initial velocity, or with external force.

Force diagram at equilibrium position: mg = F s.

Hooke’s law: Spring force F_s = −kL, where L =elongation and k =spring constant.

So: we have mg = kL which give

which gives a way to obtain k by experiment: hang a mass m and measure the elongation L.

Model the motion: Let u(t) be the displacement/position of the mass at time t, assuming the origin u = 0 is at the equilibrium position, and downward is the positive direction.

Total elongation: L + u Total spring force: F_s = −k(L + u) Other forces:

• damping/resistent force:  is the damping constant, and v is the velocity
• External force applied on the mass: F (t), given function of t

Total force on the mass:

Newton’s law of motion ma =   gives

Since mg = kL, by rearranging the terms, we get

mu′′ + γu′ + ku = F

where m ia the mass, γ is the damping constant, k is the spring constant, and F is the external force.

Next we study several cases.

Case 1: Undamped free vibration (simple harmonic motion). We assume no damping (γ = 0) and no external force (F = 0). So the equation becomes

mu′′ + ku = 0.

Solve it

General solution

u(t) = c₁ cos ω₀t + c₂ sin ω₀t.

Four terminologies of this motion: frequency, period, amplitude and phase, defined below.

Frequency:  

Period:

Amplitude and phase: We need to work on this a bit. We can write

Now, define

so we have

A trick to memorize the last term formula: Consider a right triangle, with c₁ and c₂ as the sides that form the right angle. Then, the amplitude equals to the length of the hypotenuse, and the phase δ is the angle between side c₁ and the hypotenuse. Draw a graph and you will see it better.

Problems in this part often come in the form of word problems. We need to learn the skill of extracting information from the text and put them into mathematical terms.

 Example 1. A mass weighing 10 lb stretches a spring 2 in. We neglect damping. If the mass is displaced an additional 2 in, and is then set in motion with initial upward velocity of 1 ft/sec, determine the position, frequency, period, amplitude and phase of the motion.

Answer.We see this is free harmonic oscillation. The equation is

mu′′ + ku = 0

with initial conditions (Pay attention to units!)

We need find the values for m and k. We have

To find k, we see that the elongation is    By Hooke’s law, we have

kL = mg, k = mg/L = 60.

Put in these values, we get the equation

So the frequency is  and the general solution is

u(t) = c₁ cos ω₀t + c₂ sin ω₀t

We can find c₁ , c₂ by the ICs:

(Note that c₁ = u(0) and c₂ = u′ (0)/ω₀ .) Now we have the position at any time t

The four terms of the motion are

and

Case II: Damped free vibration. We assume that γ ≠ 0(> 0) and F = 0.

mu′′ + γ u′ + ku = 0

then

We see the type of root depends on the sign of the discriminant − 4km.

• If large damping,) we have two real roots, and they are both negative. The general solution is 
   Due to the large damping force, there will be no vibration in the motion. The mass will simply return to the equilibrium position exponentially. This kind of motion is called overdamped.
• If  we have double roots r₁ = r₂ = r < 0. So u = (c₁ + c₂ t)e^rt .
  Depending on the sign of c₁ , c₂ (which is determined by the ICs), the mass may cross the equilibrium point maximum once. This kind of motion is called critically damped, and this value of γ is called critical damping.
• If  small damping) we have complex roots

So the position function is

This motion is called damped oscillation. We can re-write it as

Here the term e^−λt R is the amplitude, and µ is called the quasi frequency, and the quasi period is   The graph of the solution looks like the one for complex roots with negative real part.

Summary: For all cases, since the real part of the roots are always negative, u will go to zero as t grow. This means, if there is damping, no matter how big or small, the motion will eventually come to a rest.

Example 2. A mass of 1 kg is hanging on a spring with k = 1. The mass is in a medium that exerts a viscous resistance of 6 newton when the mass has a velocity of 48 m/sec. The mass is then further stretched for another 2m, then released from rest. Find the position u(t) of the mass.

Answer. We have  So the equation for u is

Solve it

By ICs, we have u(0) = c₁ = 2, and

So the position at any time t is

Forced Vibrations

In this chapter we assume the external force is F (t) = F₀ cos ωt. (The case where F (t) = F₀ sin ωt is totally similar.) The reason for this particular choice of force will be clear later when we learn Fourier series, i.e., we represent periodic functions with the sum of a family of sin and cos functions.

Case 1: With damping.

mu′′ + γ u′ + ku = F₀ cos ωt.

Solution consists of two parts:
u(t) = uH (t) + U (t), uH (t): the solution of the homogeneous equation, U (t): a particular solution for the non-homogeneous equation.

From discussion is the previous chapter, we know that uH → 0 as t → +∞ for systems with damping. Therefore, this part of the solution is called the transient solution.

The appearance of U is due to the force term F . Therefore it is called the forced response.

The form of this particular solution is U (t) = A₁ cos ωt + A₂ sin ωt. As we have seen, we can rewrite it as U (t) = R cos(ωt − δ) where R is the amplitude and δ is the phase.

We see it is a periodic oscillation for all time t.

As time t → ∞, we have u(t) → U (t). So U (t) is called the steady state.

Case 2: Without damping. The equation now is

mu′′ + ku = F₀ cos ωt

Let w₀ =  denote the system frequency (i.e., the frequency for the free oscillation). The homogeneous solution is

uH (t) = c₁ cos ω₀t + c₂ sin ω₀t.

The form of the particular solution depends on the value of w. We have two cases.

Case 2A: if w ≠ w₀ . The particular solution is of the form

U = A cos wt.

(Note that we did not take the sin wt term, because there is no u′ term in the equation.) And U ′′ = −w²A cos wt. Plug these in the equation

m(−w²A cos wt) + kA cos wt = F₀ cos wt,

Note that if w is close to w₀ , then A takes a large value.

General solution u(t) = c₁ cos w₀ t + c₂ sin w₀t + A cos wt where c₁ , c₂ will be determined by ICs.

Now, assume ICs:

u(0) = 0, u′(0) = 0.

Let’s find c₁ , c₂ and the solution:

u(0) = 0 : c₁ + A = 0, c₁ = −A u′ (0) = 0 : 0 + w₀c₂ + 0 = 0, c₂ = 0

Solution 

u(t) = −A cos w₀t + A cos wt = A(cos wt − cos w₀t).

We see that the solution consists of the sum of two cosine functions, with different frequencies. In order to have a better idea of how the solution looks like, we apply some manipulation.

Recall the trig identity:

We now have

Since both w0 , w are positive, then w0 + w has larger value than w0 − w. Then, the first term   can be viewed as the varying amplitude, and the second term 

 is the vibration/oscillation.

One particular situation of interests: if w₀ ≠ w but they are very close wo ≈ w, then we have |w₀ − w| << |w₀ + w|, meaning that |w₀ − w| is much smaller than |w₀ + w|. The plot of u(t) looks like (we choose w₀ = 9, w = 10)

This is called a beat. (One observes it by hitting a key on a piano that’s not tuned, for example.)

Case 2B: If w = w₀ . The particular solution is U = At cos w₀t + B t sin w₀t

A typical plot looks like:

This is called resonance. If the frequency of the source term ω equals to the frequency of the system ω₀, then, small source term could make the solution grow very large!
One can bring down a building or bridge by small periodic perturbations.

Historical disasters such as the French troop marching over a bridge and the bridge collapsed. Why? Unfortunate for the French, the system frequency of the bridge matches the frequency of their foot-steps.